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प्रश्न
If `cos^-1( x/a) +cos^-1 (y/b)=alpha` , prove that `x^2/a^2-2(xy)/(ab) cos alpha +y^2/b^2=sin^2alpha`
उत्तर
`cos^-1( x/a) +cos^-1 (y/b)=alpha`
`cos^-1(x/a) =alpha-cos^-1 (y/b)`
`=>cos {cos^-1 (x/a)}=cos{alpha-cos^-1 (y/b)}`
`=>x/a=cos alpha cos{cos^-1 (y/b)}+sinalpha sin{cos^-1(y/b)}`
`=>x/a=y/b cos alpha+sin alpha sqrt(1-(y/b)^2)`
`=>x/a - y/b cos alpha =sin alpha sqrt(1-(y/b)^2)`
Squaring both sides, we get
`(x/a - y/b cos alpha)^2={sin alpha sqrt(1-(y/b)^2)}^2`
`(x/a)^2+(y/b)^2cos^2 alpha- (2xy)/(ab) cos alpha=sin^2 alpha- sin^2 alpha(y/b)^2`
`(x/a)^2+(y/b)^2cos^2 alpha+sin^2 alpha(y/b)^2-(2xy)/(ab) cos alpha=sin^2 alpha`
`(x/a)^2+(y/b)^2(cos^2 alpha+sin^2 alpha)-(2xy)/(ab) cos alpha=sin^2 alpha`
`=>(x/a)^2-(2xy)/(ab) cos alpha+(y/b)^2=sin^2 alpha`
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